Question

If the concentration of water vapour in the air is \(1 \%\) and the total atmospheric pressure equals to 1 atm then the partial pressure of water vapour is (1) \(0.1 \mathrm{~atm}\) (2) \(1 \mathrm{~mm}\) (3) \(7.6 \mathrm{~mm}\) IIg (4) \(100 \mathrm{~atm}\)

Short answer

The partial pressure of water vapour is 7.6 mm Hg. Correct option: (3).

Step-by-step solution

- Understand the problem

The goal is to find the partial pressure of water vapour given that it constitutes 1% of the total atmospheric pressure of 1 atm.

- Recall the formula for partial pressure

The partial pressure of a gas in a mixture is given by its fraction of the total pressure. Mathematically, this can be written as: \[ P_{\text{gas}} = x_{\text{gas}} \times P_{\text{total}} \] where \( x_{\text{gas}} \) is the mole fraction (or percentage in decimal form) and \( P_{\text{total}} \) is the total pressure.

- Convert percentage to decimal

The concentration of water vapour is given as 1%. In decimal form, this is 0.01.

- Apply the formula

Using the formula from Step 2 and the decimal value from Step 3, \[ P_{\text{water vapour}} = 0.01 \times 1 \text{ atm} = 0.01 \text{ atm} \]

- Convert the partial pressure to other units

If the partial pressure needs to be expressed in mm Hg: Since 1 atm = 760 mm Hg, \[ 0.01 \text{ atm} \times 760 \text{ mm Hg/atm} = 7.6 \text{ mm Hg} \]

- Determine the correct answer

From the calculated value, the correct option is (3) 7.6 mm Hg.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas. The equation is written as \( PV = nRT \), where \( R \) is the ideal gas constant. While the ideal gas law is essential for many gas calculations, in this exercise, we focus on partial pressure, which is a different but related concept.
The ideal gas law helps us understand the behavior of gases under different conditions. In our exercise, we're considering the partial pressure of water vapor in the air. While we don't need to use the entire ideal gas law equation here, it's good to note that gases in a mixture behave ideally – meaning they follow the principles laid out in this law.

Mole Fraction

Mole fraction measures the ratio of moles of a component to the total moles in a mixture. For gases, it also reflects their proportion by volume. In our exercise, we're given a 1% concentration of water vapor.
This percentage can be directly converted to a mole fraction by dividing by 100: 1% becomes 0.01. The mole fraction lets us calculate the partial pressure of the gas. If you remember the formula for partial pressure, \( P_{\text{gas}} = x_{\text{gas}} \times P_{\text{total}} \), you can see how we use the mole fraction (0.01) and the total pressure (1 atm) to find the partial pressure of water vapor.

Unit Conversion

Unit conversion is crucial when working with different measurements. Our exercise involves atmospheric pressure given in atm (atmospheres) but also requires conversion to mm Hg (millimeters of mercury).
We use unit conversions to make our calculations relevant to the given options. In this exercise, 1 atm is known to be equal to 760 mm Hg. Therefore, for our partial pressure, converting 0.01 atm to mm Hg involves multiplying by 760 mm Hg/atm: \( 0.01 \text{ atm} \times 760 \text{ mm Hg/atm} = 7.6 \text{ mm Hg} \).
This simple operation ensures we get a result consistent with the multiple-choice format, and understanding unit conversions broadens your skillset for different chemistry problems.